Normal DTFT table contains:
$$ \cos(\omega_0 n) \xrightarrow{DTFT\ 2\pi} \pi \delta[\omega - \omega_0] + \pi \delta[\omega + \omega_0] $$
$$ \sin(\omega_0 n) \xrightarrow{DTFT\ 2\pi} i\ \pi\ \delta[\omega - \omega_0] - i\ \pi \ \delta[\omega + \omega_0] $$
How would I apply "duality property" to obtain DTFT transform for:
$$ ??? \xrightarrow{DTFT\ 2\pi} \cos(\omega_0\ \omega) $$
$$ ??? \xrightarrow{DTFT\ 2\pi} \sin(\omega_0\ \omega) $$
??? = fill in the blanks.
The answer is that you don't need to use duality to derive DTFT of sin and cos in reverse. You just need to use these identities:
$$ cos(\omega) = \frac{1}{2} (e^{i\ \omega}+e^{-i\ \omega}) $$
$$ sin(\omega) = \frac{1}{2i} (e^{i\ \omega}-e^{-i\ \omega}) $$
Then the following DTFT Transforms:
$$ \delta[n-d] \xrightarrow{DTFT} e^{-i\ \omega\ d} $$